Then I simply sub in k = 0, 1, .., 5 for all my roots. But the original equation is a polynomial of degree 3. There should be only 3 factors. Do I have to test them all to see if they work? Or is there an easier way...

then substituting each of these back, I get two 3rd degree equations which again would give me 6 roots???

That's like saying solving x^3+sqrt(1)=0 means solving x^3+1=0 and x^3-1=0. Sure you get 6 roots. The 'sqrt' notation in your notation indicates only one of them. You want the 'principal value' of the square root. It's the pi/8 one.